Find the volume of the solid whose base is the region bounded by the function f ( x ) = 9 − x 2 f\left(x\right)=\sqrt{9-x^2} and the x-axis with square cross sections perpendicular to the x-axis.

In this problem, we're asked to find the volume of the solid whose base is the region bounded by the function f of x equals nine minus x squared and the x axis with square cross sections perpendicular to the x axis. Now in this particular problem, we are not given a visual to start out. We need to come up with this visual ourselves, and this can often be the trickiest part of setting up these problems. So let's start by drawing a tilted axis here. So I'm gonna go ahead and draw my x and my y axis just sort of tilted like this. And then I have my function f of x equals nine minus x squared. So I know that this particular function represents a semicircle. So if I go ahead and draw this semicircle, the square root of nine minus X squared, both of the endpoints on my x axis here are going to be negative three and positive three because that's where my function is equal to zero. Now we're told that the base of this three-dimensional shape is the region bounded by this function and the x axis. So if I go ahead and draw another line around my or along my x axis here, this represents the base of my solid. So I'm actually going to go ahead and shade it in in this sort of orange color. So this represents the base of my solid, and we're taking this solid and we are extending or we're taking this base and we're extending it upwards to form our solid. So if I take this shape, I sort of extend it upwards. Then I have this three-dimensional shape that we're working with here. Now we're told here that there are square cross sections when we cut perpendicular to the x axis. Now what does it mean to be perpendicular to the x axis? It just means that we're slicing it like this. So if I slice through this shape, I end up getting a square cross section. So the cross section of this solid is a square. So I'm going to go ahead and draw my front facing cross section here as a square. So what are the side lengths of my square? Because ultimately, we wanna find the volume of this shape. And having these square cross sections is going to help us do that because to find our volume, we are integrating the area function of a cross section. If my cross section is a square, then I need to know what these side lengths are to find this area. Now looking at this, I can see that that bottom side length is between my function F of X and the X axis. So that ultimately means that this side length is F of X. Now because this is a square, that means all of my side lengths are F of X. So I have my side length F of X. I have my other side length F of X, which means that the area of this cross section is f of x squared. Now this specifically happens when we're dealing with the square cross sections. So with that area in mind, let's set up our volume integral. The volume of this shape is going to be equal to the definite integral. So here we can see that it's bounded from negative three to three because, again, that's where our function is equal to zero. So negative three to three of f of x, which here is the square root of nine minus X squared squared. So we're just plugging in that area function into our volume integral and we're integrating with respect to X. So now we just need to evaluate this integral now that we have it set up. I can see that this square and this square root nicely cancel out here. So this is just from negative three to three of nine minus X squared DX. Now applying my fundamental theorem of calculus here, I get nine X minus one third X cubed, bounded from negative three to three, and then I can just plug in those bounds. So I get nine times three minus one third three cubed and that that's my upper bound, and then subtracting off my lower bound here, that's going to be nine times negative three minuteus one third times negative three cubed. So nine times three is gonna be 27, so that's 27, and then three cubed is also 27, divide that by three we get nine, so 27 minus nine, and then minus nine times negative three, that's going to be negative 27, and then negative three cubed that is also negative 27. Divide that by three, that is nine, but this becomes positive because we have that double negative. 27 minus nine is 18. Negative 27 plus nine is negative 18. So we have minus negative 18. 18 minuteus negative 18, that's 18 plus 18. This ultimately gives us here thirty thirty six. And this 36 isn't just a random number. It represents the volume of this three-dimensional shape described us in our problem. Now if you have any questions here either about how I set up this integral or any of the algebra that I did along the way, let us know in the comments. I'll see you in the next one.

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Calculus

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

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객관식

Find the volume of the solid whose base is the region bounded by the function $f (x) = \sqrt{9 - x^{2}}$ and the x-axis with square cross sections perpendicular to the x-axis.

검증된 단계별 안내

Identify the region bounded by the function f(x) = $\sqrt{9-x^2}$ and the x-axis. This is a semicircle with radius 3 centered at the origin.

Determine the limits of integration. Since the semicircle is centered at the origin and has a radius of 3, the limits of integration are from x = -3 to x = 3.

Recognize that the cross sections perpendicular to the x-axis are squares. The side length of each square is given by the function f(x) = $\sqrt{9-x^2}$.

Calculate the area of each square cross section, A(x), as a function of x. Since the side length is f(x), the area is A(x) = ($\sqrt{9-x^2}$)^2 = 9-x^2.

Set up the integral to find the volume of the solid: $\int$_{-3}^{3} (9-x^2) \, dx. Evaluate this integral to find the volume of the solid.

4:53m

다음 시청

마스터 Finding Volume Using Disks Example 3가 Nick Kaneko 짧은 영상으로 설명해 줍니다

Introduction to Volume & Disk Method 연습 세트

수석 튜터가 만든 문제 세트

전문가 영상 설명

Calculus 때문에 힘들어하고 계신가요?

Find the volume of the solid whose base is the region bounded by the function f(x)=9−x2f\(\left\)(x\(\right\))=\(\sqrt{9-x^2}\) and the x-axis with square cross sections perpendicular to the x-axis.

다음 시청

Introduction to Volume & Disk Method 연습 세트

Find the volume of the solid whose base is the region bounded by the function $f (x) = \sqrt{9 - x^{2}}$ and the x-axis with square cross sections perpendicular to the x-axis.